EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 6013 ISSN 1307-5543 – ejpam.com Published by New York Business Global On R-(τ1, τ2)-continuous Functions Napassanan Srisarakham1, Supunnee Sompong2, Chawalit Boonpok1,∗ 1 Mathematics and Applied Mathematics Research Unit, Department of Mathematics, Faculty of Science, Mahasarakham University, Maha Sarakham, 44150, Thailand 2 Department of Mathematics and Statistics, Faculty of Science and Technology, Sakon Nakhon Rajbhat University, Sakon Nakhon, 47000, Thailand Abstract. This paper presents a new class of functions between bitopological spaces called R- (τ1, τ2)-continuous functions. Furthermore, several characterizations and some properties concern- ing R-(τ1, τ2)-continuous functions are established. 2020 Mathematics Subject Classifications: 54C08, 54E55 Key Words and Phrases: τ1τ2-open set, R-(τ1, τ2)-continuous function 1. Introduction The field of the mathematical science which goes under the name of topology is con- cerned with all questions directly or indirectly related to continuity. Preopen sets, semi- open sets, α-open sets, β-open sets, δ-open sets and θ-open sets play an important role in the research of generalizations of continuity. By using these sets many authors introduced and investigated various types of continuity. In [1], the present authors studied some prop- erties of (Λ, sp)-open sets, s(Λ, sp)-open sets, p(Λ, sp)-open sets, α(Λ, sp)-open sets and β(Λ, sp)-open sets. Viriyapong and Boonpok [2] investigated several characterizations of (Λ, sp)-continuous functions by utilizing the notions of (Λ, sp)-open sets and (Λ, sp)-closed sets. Dungthaisong et al. [3] introduced and studied the concept of g(m,n)-continuous functions. Duangphui et al. [4] introduced and investigated the notion of (µ, µ′)(m,n)- continuous functions. Moreover, some characterizations of almost (Λ, p)-continuous func- tions, almost strongly θ(Λ, p)-continuous functions, weakly (Λ, b)-continuous functions, θ(⋆)-precontinuous functions, (Λ, p(⋆))-continuous functions, ⋆-continuous functions, θ-I - continuous functions, almost (g,m)-continuous functions, pairwise almost M -continuous functions were presented in [5], [6], [7], [8], [9], [10], [11], [12] and [13], respectively. Konstadilaki-Savvopoulou and Janković [14] introduced and studied a strong form of conti- nuity of functions between topological spaces called R-continuous functions. Crossley and ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6013 Email addresses: napassanan.sri@msu.ac.th (N. Srisarakham), s−sompong@snru.ac.th (S. Sompong), chawalit.b@msu.ac.th (C. Boonpok) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 2 of 12 Hildebrand [15] introduced and investigated the concept of irresolute functions. Reilly and Vamanamurthy [16] introduced and studied the notion of preirresolute functions. Baker [17] introduced and investigated the concept of R-irrsolute functions. Further- more, the present author [18] introduced a strong form of preirresolute functions called R-preirresolute functions. These four classes of functions have properties similar to the class of R-continuous functions. Beceren and Noiri [19] introduced and studied the notions of new classes of functions, namely α-preirresolute functions and β-preirresolute functions. A new class of α-preirresolute functions which is stronger than preirresolute functions [17] is a generalization of strongly M -precontinuous functions [20]. A new class of β-irresolute functions which is stronger than almost α-irresolute functions [19] is a generalization of preirresolute functions [17]. Noiri and Popa [21] introduced a new class of functions called R-M -continuous functions as functions defined between sets satisfying some minimal con- ditions and obtained several characterizations of R-M -continuous functions. Noiri and Popa [21] investigated the relationship between R-M -continuity and some low separation axioms (m-T1, m-T2, m-R0). The notion of (τ1, τ2)-continuous functions was introduced in [22]. Moreover, several characterizations of almost (τ1, τ2)-continuous functions and weakly (τ1, τ2)-continuous functions were studied in [23] and [24], respectively. In this paper, we introduce the concept of R-(τ1, τ2)-continuous functions. We also investigate some characterizations of R-(τ1, τ2)-continuous functions. 2. Preliminaries Throughout the present paper, spaces (X, τ1, τ2) and (Y, σ1, σ2) (or simply X and Y ) always mean bitopological spaces on which no separation axioms are assumed unless explicitly stated. Let A be a subset of a bitopological space (X, τ1, τ2). The closure of A and the interior of A with respect to τi are denoted by τi-Cl(A) and τi-Int(A), respectively, for i = 1, 2. A subset A of a bitopological space (X, τ1, τ2) is called τ1τ2-closed [25] if A = τ1-Cl(τ2-Cl(A)). The complement of a τ1τ2-closed set is called τ1τ2-open. The intersection of all τ1τ2-closed sets of X containing A is called the τ1τ2-closure [25] of A and is denoted by τ1τ2-Cl(A). The union of all τ1τ2-open sets of X contained in A is called the τ1τ2-interior [25] of A and is denoted by τ1τ2-Int(A). Lemma 1. [25] Let A and B be subsets of a bitopological space (X, τ1, τ2). For the τ1τ2- closure, the following properties hold: (1) A ⊆ τ1τ2-Cl(A) and τ1τ2-Cl(τ1τ2-Cl(A)) = τ1τ2-Cl(A). (2) If A ⊆ B, then τ1τ2-Cl(A) ⊆ τ1τ2-Cl(B). (3) τ1τ2-Cl(A) is τ1τ2-closed. (4) A is τ1τ2-closed if and only if A = τ1τ2-Cl(A). (5) τ1τ2-Cl(X −A) = X − τ1τ2-Int(A). N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 3 of 12 A subset A of a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)r-open [26] (resp. (τ1, τ2)s-open [27], (τ1, τ2)p-open [27], (τ1, τ2)β-open [27]) if A = τ1τ2-Int(τ1τ2-Cl(A)) (resp. A ⊆ τ1τ2-Cl(τ1τ2-Int(A)), A ⊆ τ1τ2-Int(τ1τ2-Cl(A)), A ⊆ τ1τ2-Cl(τ1τ2-Int(τ1τ2-Cl(A)))). The complement of a (τ1, τ2)r-open (resp. (τ1, τ2)s-open, (τ1, τ2)p-open, (τ1, τ2)β-open) set is called (τ1, τ2)r-closed (resp. (τ1, τ2)s-closed, (τ1, τ2)p-closed, (τ1, τ2)β-closed). A subset A of a bitopological space (X, τ1, τ2) is said to be α(τ1, τ2)-open [28] if A ⊆ τ1τ2-Int(τ1τ2-Cl(τ1τ2-Int(A))). The complement of an α(τ1, τ2)-open set is said to be α(τ1, τ2)-closed. Let A be a subset of a bitopological space (X, τ1, τ2). A point x ∈ X is called a (τ1, τ2)θ-cluster point [26] of A if τ1τ2-Cl(U) ∩ A ̸= ∅ for every τ1τ2-open set U containing x. The set of all (τ1, τ2)θ-cluster points of A is called the (τ1, τ2)θ-closure [26] of A and is denoted by (τ1, τ2)θ-Cl(A). A subset A of a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)θ-closed [26] if A = (τ1, τ2)θ-Cl(A). The complement of a (τ1, τ2)θ-closed set is said to be (τ1, τ2)θ-open. The union of all (τ1, τ2)θ-open sets contained in A is called the (τ1, τ2)θ-interior [26] of A and is denoted by (τ1, τ2)θ-Int(A). Lemma 2. [26] For a subset A of a bitopological space (X, τ1, τ2), the following properties hold: (1) If A is τ2τ2-open in X, then τ1τ2-Cl(A) = (τ1, τ2)θ-Cl(A). (2) (τ1, τ2)θ-Cl(A) is τ1τ2-closed in X. 3. R-(τ1, τ2)-continuous functions In this section, we introduce the concept of R-(τ1, τ2)-continuous functions. Further- more, several characterizations of R-(τ1, τ2)-continuous functions are discussed. Definition 1. A functions f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be R-(τ1, τ2)-continuous if for each x ∈ X and for each σ1σ2-open set V of Y containing f(x), there exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(U)) ⊆ V . Definition 2. [22] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called (τ1, τ2)-continuous at a point x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a τ1τ2-open set U of X containing x such that f(U) ⊆ V . A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called (τ1, τ2)-continuous if f has this property at each point of X. Lemma 3. [22] For a function (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equiv- alent: (1) f is (τ1, τ2)-continuous; (2) f−1(V ) is τ1τ2-open in X for every σ1σ2-open set V of Y ; (3) f(τ1τ2-Cl(A)) ⊆ σ1σ2-Cl(f(A)) for every subset A of X; (4) τ1τ2-Cl(f −1(B)) ⊆ f−1(σ1σ2-Cl(B)) for every subset B of Y ; N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 4 of 12 (5) f−1(σ1σ2-Int(B)) ⊆ τ1τ2-Int(f −1(B)) for every subset B of Y ; (6) f−1(K) is τ1τ2-closed in X for every σ1σ2-closed set K of Y . Lemma 4. If a function f : (X, τ1, τ2) → (Y, σ1, σ2) is R-(τ1, τ2)-continuous, then f is (τ1, τ2)-continuous. Proof. Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since f is R-(τ1, τ2)-continuous, there exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(U)) ⊆ V . This implies that f(U) ⊆ V . Thus, f is (τ1, τ2)-continuous. Theorem 1. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is R-(τ1, τ2)-continuous; (2) for each point x ∈ X and each σ1σ2-open set V of Y containing f(x), exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(τ1τ2-Cl(U))) ⊆ V ; (3) for each point x ∈ X and each σ1σ2-closed set F of Y with f(x) ̸∈ F , exists a τ1τ2- open set U of X containing x and a σ1σ2-open set V of Y such that F ⊆ V and f(τ1τ2-Cl(U)) ∩ V = ∅; (4) for each point x ∈ X and each σ1σ2-closed set F of Y with f(x) ̸∈ F , exists a τ1τ2- open set U of X containing x and a σ1σ2-open set V of Y such that F ⊆ V and f(U) ∩ V = ∅. Proof. (1) ⇒ (2): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since f is R-(τ1, τ2)-continuous, there exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(U)) ⊆ V . By Lemma 4, we have f is (τ1, τ2)-continuous and by Lemma 3, f(τ1τ2-Cl(U)) ⊆ σ1σ2-Cl(f(U)) ⊆ V . Thus, σ1σ2-Cl(f(τ1τ2-Cl(U))) ⊆ σ1σ2-Cl(f(U)) ⊆ V. (2) ⇒ (3): Let x ∈ X and F be any σ1σ2-closed set of Y with f(x) ̸∈ F . Then, we have f(x) ∈ Y − F and Y − F is σ1σ2-open in Y . By (2), there exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(τ1τ2-Cl(U))) ⊆ Y − F . Put V = Y − σ1σ2-Cl(f(τ1τ2-Cl(U))). Then, V is σ1σ2-open and F ⊆ V . Furthermore, f(τ1τ2-Cl(U)) ∩ V = ∅. (3) ⇒ (4): The proof is obvious. (4) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Then, Y − V is σ1σ2-closed in Y and f(x) ̸∈ Y − V . By (4), there exists a τ1τ2-open set U of X containing x and a σ1σ2-open set W of Y such that Y −V ⊆ W and f(U)∩W = ∅. Since f(U) ⊆ Y −W and Y −W is σ1σ2-closed, σ1σ2-Cl(f(U)) ⊆ σ1σ2-Cl(Y −W ) = Y −W ⊆ V . This shows that f is R-(τ1, τ2)-continuous. N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 5 of 12 Definition 3. [29] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be strongly θ(τ1, τ2)- continuous at a point x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a τ1τ2-open set U of X containing x such that f(τ1τ2-Cl(U)) ⊆ V . A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be strongly θ(τ1, τ2)-continuous if f is strongly θ(τ1, τ2)-continuous at each point x of X. Theorem 2. If a function f : (X, τ1, τ2) → (Y, σ1, σ2) is R-(τ1, τ2)-continuous, then f is strongly θ(τ1, τ2)-continuous. Proof. It follows from Theorem 1(2). Definition 4. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be weakly (τ1, τ2)-closed if for each τ1τ2-closed set F of X, σ1σ2-Cl(f(τ1τ2-Int(F ))) ⊆ f(F ). Theorem 3. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is weakly (τ1, τ2)-closed; (2) σ1σ2-Cl(f(U)) ⊆ f(τ1τ2-Cl(U)) for each τ1τ2-open set U of X; (3) for each subset B of Y and each τ1τ2-open set U of X with f−1(B) ⊆ U , there exists a σ1σ2-open set V of Y such that B ⊆ V and f−1(V ) ⊆ τ1τ2-Cl(U); (4) for each y ∈ Y and each τ1τ2-open set U of X with f−1(y) ⊆ U , there exists a σ1σ2-open set V of Y containing y such that f−1(y) ⊆ τ1τ2-Cl(U); (5) σ1σ2-Cl(f(τ1τ2-Int(τ1τ2-Cl(A)))) ⊆ f(τ1τ2-Cl(A)) for each subset A of X; (6) σ1σ2-Cl(f(τ1τ2-Int((τ1, τ2)θ-Cl(A)))) ⊆ f((τ1, τ2)θ-Cl(A)) for each subset A of X. Proof. (1) ⇒ (2): Let U be any τ1τ2-open set of X. Since f is weakly (τ1, τ2)-closed, we have σ1σ2-Cl(f(U)) ⊆ σ1σ2-Cl(f(τ1τ2-Int(τ1τ2-Cl(U)))) ⊆ f(τ1τ2-Cl(U)). (2) ⇒ (3): Let B be any subset of Y and U be any τ1τ2-open set ofX with f−1(B) ⊆ U . Since U is τ1τ2-open, f−1(B) ∩ τ1τ2-Cl(X − τ1τ2-Cl(U)) = f−1(B) ∩ (X − τ1τ2-Int(τ1τ2-Cl(U))) ⊆ f−1(B) ∩ (X − τ1τ2-Int(U)) = f−1(B) ∩ (X − U) = ∅. Thus, f−1(B) ∩ τ1τ2-Cl(X − τ1τ2-Cl(U)) = ∅ and hence B ∩ f(τ1τ2-Cl(X − τ1τ2-Cl(U))) = ∅. On the other hand, we have τ1τ2-Cl(U) is τ1τ2-closed and X − τ1τ2-Cl(U) is τ1τ2-open. Thus by (2), σ1σ2-Cl(f(X − τ1τ2-Cl(U))) ⊆ f(τ1τ2-Cl(X − τ1τ2-Cl(U))) and so B ∩ σ1σ2-Cl(f(X − τ1τ2-Cl(U))) ⊆ B ∩ f(τ1τ2-Cl(X − τ1τ2-Cl(U))). N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 6 of 12 Let V = Y − σ1σ2-Cl(f(X − τ1τ2-Cl(U))). Then, V is σ1σ2-open in Y , B ⊆ V and f−1(V ) = f−1(Y − σ1σ2-Cl(f(X − τ1τ2-Cl(U)))) ⊆ X − f−1(f(X − τ1τ2-Cl(U))) ⊆ τ1τ2-Cl(U). (3) ⇒ (4): The proof is obvious. (4) ⇒ (1): Let F be any τ1τ2-closed set of X and y ∈ Y − f(F ). Since Y − F is σ1σ2-open and f−1(y) ⊆ X − F , by (4) there exists a σ1σ2-open set V of Y with y ∈ V and f−1(V ) ⊆ τ1τ2-Cl(X − F ) = X − τ1τ2-Int(F ). Thus, V ∩ f(τ1τ2-Int(F )) = ∅ and hence y ∈ Y −σ1σ2-Cl(f(τ1τ2-Int(F ))). Therefore, σ1σ2-Cl(f(τ1τ2-Int(F ))) ⊆ f(F ) which implies that f is weakly (τ1, τ2)-closed. (1) ⇒ (5): Let A be any subset of X. Then, τ1τ2-Cl(A) is τ1τ2-closed and by (1), σ1σ2-Cl(f(τ1τ2-Int(τ1τ2-Cl(A)))) ⊆ f(τ1τ2-Cl(A)). (5) ⇒ (2): Let U be any τ1τ2-open set of X. Then by (5), σ1σ2-Cl(f(U)) ⊆ σ1σ2-Cl(f(τ1τ2-Int(τ1τ2-Cl(U)))) ⊆ f(τ1τ2-Cl(U)). (1) ⇒ (6): Let A be any subset of X. Thus by Lemma 2, (τ1, τ2)θ-Cl(A) is τ1τ2-closed and by (1), we have σ1σ2-Cl(f(τ1τ2-Int((τ1, τ2)θ-Cl(A)))) ⊆ f((τ1, τ2)θ-Cl(A)). (6) ⇒ (2): Let U be any τ1τ2-open set of X. By Lemma 2, we have τ1τ2-Cl(U) = (τ1, τ2)θ-Cl(U) and by (6), σ1σ2-Cl(f(U)) ⊆ σ1σ2-Cl(f(τ1τ2-Int((τ1, τ2)θ-Cl(U)))) ⊆ f((τ1, τ2)θ-Cl(U)) = f(τ1τ2-Cl(U)). Theorem 4. If a function f : (X, τ1, τ2) → (Y, σ1, σ2) is strongly θ(τ1, τ2)-continuous and weakly (τ1, τ2)-closed, then f is R-(τ1, τ2)-continuous. Proof. Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since f is strongly θ(τ1, τ2)-continuous, there exists a τ1τ2-open set U of X containing x such that f(τ1τ2-Cl(U)) ⊆ V . Since f is weakly (τ1, τ2)-closed, by Theorem 3 we have σ1σ2-Cl(f(U)) ⊆ f(τ1τ2-Cl(U)) ⊆ V. This shows that f is R-(τ1, τ2)-continuous. Definition 5. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be contra-(τ1, τ2)-open if f(U) is σ1σ2-closed in Y for every τ1τ2-open set U of X. N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 7 of 12 Theorem 5. If a function f : (X, τ1, τ2) → (Y, σ1, σ2) is contra-(τ1, τ2)-open, then f is weakly (τ1, τ2)-closed. Proof. Let F be any τ1τ2-closed set of X. Then, τ1τ2-Int(F ) is a τ1τ2-open set of X. Since f is contra-(τ1, τ2)-open, we have f(τ1τ2-Int(F )) is σ1σ2-closed in Y and hence σ1σ2-Cl(f(τ1τ2-Int(F ))) = f(τ1τ2-Int(F )) ⊆ f(F ). This shows that f is weakly (τ1, τ2)- closed. Theorem 6. If a function f : (X, τ1, τ2) → (Y, σ1, σ2) is (τ1, τ2)-continuous and contra- (τ1, τ2)-open, then f is R-(τ1, τ2)-continuous. Proof. Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since f is (τ1, τ2)-continuous, by Lemma 3 we have f−1(V ) is τ1τ2-open in X. Since f is contra- (τ1, τ2)-open, f(f −1(V )) is σ1σ2-closed in Y and σ1σ2-Cl(f(f −1(V ))) = f(f−1(V )) ⊆ V . Put U = f−1(V ). Then, U is a τ1τ2-open set of X containing x and σ1σ2-Cl(f(U)) ⊆ V . This shows that f is R-(τ1, τ2)-continuous. Definition 6. [24] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be weakly (τ1, τ2)- continuous at a point x ∈ X if for each σ1σ2-open set V of Y containing f(x), there exists a τ1τ2-open set U of X containing x such that f(U) ⊆ σ1σ2-Cl(V ). A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to be weakly (τ1, τ2)-continuous if f has this property at each point of X. Lemma 5. [24] For a function (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equiv- alent: (1) f is weakly (τ1, τ2)-continuous; (2) f(τ1τ2-Cl(A)) ⊆ (σ1, σ2)θ-Cl(f(A)) for every subset A of X; (3) τ1τ2-Cl(f −1(B)) ⊆ f−1((σ1, σ2)θ-Cl(B)) for every subset B of Y . Recall that a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)-regular [30] if for each τ1τ2-closed set F and each x ̸∈ F , there exist disjoint τ1τ2-open sets U and V such that x ∈ U and F ⊆ V . Lemma 6. [31] A bitopological space (X, τ1, τ2) is (τ1, τ2)-regular if and only if for each x ∈ X and each τ1τ2-open set U containing x, there exists a τ1τ2-open set V such that x ∈ V ⊆ τ1τ2-Cl(V ) ⊆ U . Lemma 7. [31] Let (X, τ1, τ2) be a (τ1, τ2)-regular space. Then, the following properties hold: (1) τ1τ2-Cl(A) = (τ1, τ2)θ-Cl(A) for every subset A of X. (2) Every τ1τ2-open set is (τ1, τ2)θ-open. N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 8 of 12 Theorem 7. If f : (X, τ1, τ2) → (Y, σ1, σ2) is weakly (τ1, τ2)-continuous and (Y, σ1, σ2) is (σ1, σ2)-regular, then f is R-(τ1, τ2)-continuous. Proof. Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since (Y, σ1, σ2) is (σ1, σ2)-regular, by Lemma 6 there exists a σ1σ2-open set W of Y such that f(x) ∈ W ⊆ σ1σ2-Cl(W ) ⊆ V. Since f is weakly (τ1, τ2)-continuous, there exists a τ1τ2-open set U of X containing x such that f(U) ⊆ σ1σ2-Cl(W ). Thus, σ1σ2-Cl(f(U)) ⊆ σ1σ2-Cl(W ) ⊆ V . This shows that f is R-(τ1, τ2)-continuous. Definition 7. [32] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called faintly (τ1, τ2)-continuous at a point x ∈ X if for each (σ1, σ2)θ-open set V of Y containing f(x), there exists a τ1τ2- open set U of X containing x such that f(U) ⊆ V . A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called faintly (τ1, τ2)-continuous if f has this property at every point of X. Lemma 8. [32] For a function f : (X, τ1, τ2) → (Y, σ1, σ2), the following properties are equivalent: (1) f is faintly (τ1, τ2)-continuous; (2) f−1(V ) is τ1τ2-open in X for each (σ1, σ2)θ-open set V of Y ; (3) f−1(K) is τ1τ2-closed in X for each (σ1, σ2)θ-closed set K of Y . Theorem 8. For a function f : (X, τ1, τ2) → (Y, σ1, σ2), where (Y, σ1, σ2) is (σ1, σ2)- regular, the following properties are equivalent: (1) f is R-(τ1, τ2)-continuous; (2) f is strongly θ(τ1, τ2)-continuous; (3) f is (τ1, τ2)-continuous; (4) f is weakly (τ1, τ2)-continuous; (5) f is faintly (τ1, τ2)-continuous. Proof. (1) ⇒ (2): It follows from Theorem 2. (2) ⇒ (3) and (3) ⇒ (4): The proofs are obvious. (4) ⇒ (5): Let F be any θ(τ1, τ2)-closed set of Y . Since f is weakly (τ1, τ2)-continuous, by Lemma 5 we have σ1σ2-Cl(f −1(F )) ⊆ f−1((σ1, σ2)θ-Cl(F )) = f−1(F ) and hence f−1(F ) is τ1τ2-closed in X. Thus by Lemma 8, f is faintly (τ1, τ2)-continuous. (5) ⇒ (1): Let x ∈ X and V be any σ1σ2-open set of Y containing f(x). Since (Y, σ1, σ2) is (σ1, σ2)-regular, by Lemma 7 we have V is a θ(τ1, τ2)-open set of Y . Since f is faintly (τ1, τ2)-continuous, by Lemma 8 we have f−1(V ) is τ1τ2-open in X. Then by Lemma 3, f is (τ1, τ2)-continuous. N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 9 of 12 Definition 8. A function f : (X, τ1, τ2) → (Y, σ1, σ2) is called (τ1, τ2)-open if f(V ) is σ1σ2-open in Y for every τ1τ2-open set V of X. Theorem 9. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a R-(τ1, τ2)-continuous and (τ1, τ2)-open surjection, then (Y, σ1, σ2) is (σ1, σ2)-regular. Proof. Let y ∈ Y and V be any σ1σ2-open set of Y containing y. Let x ∈ f−1(V ). Since f is R-(τ1, τ2)-continuous, there exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(U)) ⊆ V . Since f is (τ1, τ2)-open, we have f(U) is σ1σ2-open in Y and hence y ∈ f(U) ⊆ σ1σ2-Cl(f(U)) ⊆ V . It follows from Lemma 6 that (Y, σ1, σ2) is (σ1, σ2)-regular. Corollary 1. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a (τ1, τ2)-continuous and (τ1, τ2)-open surjection, then f is R-(τ1, τ2)-continuous if and only if (Y, σ1, σ2) is (σ1, σ2)-regular. Proof. This is an immediate consequence of Theorem 7 and Theorem 9. Lemma 9. [33] Let (X, τ1, τ2) be (τ1, τ2)-regular. Then, a function f : (X, τ1, τ2) → (Y, σ1, σ2) is strongly θ(τ1, τ2)-continuous if and only if f is (τ1, τ2)-continuous. Theorem 10. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a (τ1, τ2)-continuous, (τ1, τ2)-open and weakly (τ1, τ2)-closed surjection and (X, τ1, τ2) is (τ1, τ2)-regular, then (Y, σ1, σ2) is (σ1, σ2)- regular. Proof. Since f is (τ1, τ2)-continuous and (X, τ1, τ2) is (τ1, τ2)-regular, by Lemma 9 we have f is strongly θ(τ1, τ2)-continuous. Furthermore, since f is weakly (τ1, τ2)-closed, by Theorem 4 we have f is R-(τ1, τ2)-continuous. It follows from Theorem 9 that (Y, σ1, σ2) is (σ1, σ2)-regular. Definition 9. [29] A function f : (X, τ1, τ2) → (Y, σ1, σ2) is said to have a strongly θ(τ1, τ2)-closed graph with respect to X if for each (x, y) ∈ (X × Y ) − G(f), there exist a τ1τ2-open set U of X containing x and a σ1σ2-open set V of Y containing y such that [τ1τ2-Cl(U)× V ] ∩G(f) = ∅. Lemma 10. [29] A function f : (X, τ1, τ2) → (Y, σ1, σ2) has a strongly θ(τ1, τ2)-closed graph with respect to X if and only if for each (x, y) ∈ (X × Y ) − G(f), there exist a τ1τ2-open set U of X containing x and a σ1σ2-open set V of Y containing y such that f(τ1τ2-Cl(U)) ∩ V = ∅. Recall that a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)-T1 [34] if for any pair of distinct points x, y in X, there exist τ1τ2-open sets U and V of X such that x ∈ U , y ̸∈ U and y ∈ V , x ̸∈ V . N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 10 of 12 Theorem 11. If a function f : (X, τ1, τ2) → (Y, σ1, σ2) is R-(τ1, τ2)-continuous and (Y, σ1, σ2) is (σ1, σ2)-T1, then G(f) is strongly θ(τ1, τ2)-closed with respect to X. Proof. Let (x, y) ∈ (X × Y ) − G(f). Then, y ̸= f(x). Since (Y, σ1, σ2) is (σ1, σ2)-T1, there exists a σ1σ2-open set V of Y such that f(x) ∈ V and y ̸∈ V . Since f is R-(τ1, τ2)- continuous, there exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(U)) ⊆ V . Since y ̸∈ V , we have y ̸∈ σ1σ2-Cl(f(U)) and so y ∈ Y − σ1σ2-Cl(f(U)). By Lemma 1, σ1σ2-Cl(f(U)) is σ1σ2-closed and Y − σ1σ2-Cl(f(U)) is σ1σ2-open. Since f is R-(τ1, τ2)- continuous, f is (τ1, τ2)-continuous and by Lemma 3, f(τ1τ2-Cl(U)) ⊆ σ1σ2-Cl(f(U)). Then, f(τ1τ2-Cl(U)) ∩ (Y − σ1σ2-Cl(f(U))) = ∅ and by Lemma 10, G(f) is strongly θ(τ1, τ2)-closed with respect to X. Recall that a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)-T2 [35] if for any pair of distinct points x, y in X, there exist disjoint τ1τ2-open sets U and V of X containing x and y, respectively. Theorem 12. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a (τ1, τ2)-continuous injection and (Y, σ1, σ2) is (σ1, σ2)-T2, then (X, τ1, τ2) is (τ1, τ2)-T2. Proof. Suppose that (Y, σ1, σ2) is (σ1, σ2)-T2. Let x, y be any distinct points of X. Since f is injective, f(x) ̸= f(y). Since (Y, σ1, σ2) is (σ1, σ2)-T2, there exist σ1σ2-open sets V and W of Y containing f(x) and f(y), respectively, such that V ∩W = ∅. Since f is (τ1, τ2)-continuous, there exist τ1τ2-open sets U and G of X containing x and y, respectively, such that f(U) ⊆ V and f(G) ⊆ W . It follows that U ∩ G = ∅. Thus, (X, τ1, τ2) is (τ1, τ2)-T2. Corollary 2. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a R-(τ1, τ2)-continuous injection and (Y, σ1, σ2) is (σ1, σ2)-T2, then (X, τ1, τ2) is (τ1, τ2)-T2. Proof. This is an immediate consequence of Lemma 4 and Theorem 12. Recall that a bitopological space (X, τ1, τ2) is said to be (τ1, τ2)-R0 [36] if for each τ1τ2-open set U and each x ∈ U , τ1τ2-Cl({x}) ⊆ U . Theorem 13. If f : (X, τ1, τ2) → (Y, σ1, σ2) is a R-(τ1, τ2)-continuous surjection, then (Y, σ1, σ2) is (σ1, σ2)-R0. Proof. Let V be any σ1σ2-open set of Y and y ∈ V . Let x ∈ X such that y = f(x). Since f is R-(τ1, τ2)-continuous, there exists a τ1τ2-open set U of X containing x such that σ1σ2-Cl(f(U)) ⊆ V . Thus, σ1σ2-Cl({y}) = σ1σ2-Cl({f(x)}) ⊆ σ1σ2-Cl(f(U)) ⊆ V and hence (Y, σ1, σ2) is (σ1, σ2)-R0. Acknowledgements This research project was financially supported by Mahasarakham University. N. Srisarakham, S. Sompong, C. Boonpok / Eur. J. Pure Appl. Math, 18 (2) (2025), 6013 11 of 12 References [1] C. Boonpok and J. Khampakdee. (Λ, sp)-open sets in topological spaces. European Journal of Pure and Applied Mathematics, 15(2):572–588, 2022. [2] C. Viriyapong and C. Boonpok. 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